One of the easiest ways to understand that type of problem is simply to visualize the parabola of the original inequality. Equation [tex]3x^2 + 12x[/tex] is a positive (or upward) parabola right! So just think about where, in relation to its zeros, that such a parabola sits above the x axis.

The factoring was good. The best approach is to look for critical points (the values of x for which the expression becomes 0). Check a value for x in each interval and determine the truthfulness of the relation.

You can easily keep to one dimension, a single number line graph, for this kind of exercise.

If AB> 0 it does NOT follow that A> 0 or B> 0. It might happen that A< 0 and B< 0.

A good way to solve general inequalities is to solve the associated equation first. For your example, 3x2+ 12x> 0, the associated equality is 3x2+ 12x= 3x(x+ 4)= 0 has roots x= 0, x= -4. That divides the real line into 3 intervals: [itex](-\infty, -4)[/itex], (-4, 0), and [itex](0, \infty)[/itex]. If x< -4, then both x and x+4 are negative: their product is positive. If -4< x< 0, then x is still negative but x+ 4 is positive: their product is negative. If x> 0, then both x and x+ 4 are positive: their product is positive. 3x2+ 12x> 0 is satisfied for x< -4 or x> 0.

If AB> 0 it does NOT follow that A> 0 or B> 0. It might happen that A< 0 and B< 0.

A good way to solve general inequalities is to solve the associated equation first. For your example, 3x2+ 12x> 0, the associated equality is 3x2+ 12x= 3x(x+ 4)= 0 has roots x= 0, x= -4. That divides the real line into 3 intervals: [itex](-\infty, -4)[/itex], (-4, 0), and [itex](0, \infty)[/itex]. If x< -4, then both x and x+4 are negative: their product is positive. If -4< x< 0, then x is still negative but x+ 4 is positive: their product is negative. If x> 0, then both x and x+ 4 are positive: their product is positive. 3x2+ 12x> 0 is satisfied for x< -4 or x> 0.

Why, by the way, was this titled "solving an absolute value"?

Thanks for all the replies. My mistake, I titled it that because it had a > sign xP I dunno, many absolute value problems have the signs lol. I think I need to go back and look at this. I've forgotten how to do this and I know that I've learned it last year. It makes sense to me though. Thanks for your help!